document.write( "Question 562910: A train running from A to B meets with an accident 50 miles from A, after which it travels 3/5 of its original velocity and arrives 3 hours late at B; if the accident has occurred 50 miles further on, it would have been 2 hours late. Find the distance from A to B and the original velocity of the train? \n" ); document.write( "
Algebra.Com's Answer #364939 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! A train running from A to B meets with an accident 50 miles from A, after which it travels 3/5 of its original velocity and arrives 3 hours late at B; \n" ); document.write( " if the accident has occurred 50 miles further on, it would have been 2 hours late. \n" ); document.write( "Find the distance from A to B and the original velocity of the train? \n" ); document.write( ": \n" ); document.write( "Let s = normal speed of the train \n" ); document.write( "then \n" ); document.write( ".6s = speed after accident \n" ); document.write( ": \n" ); document.write( "d = distance from A to B \n" ); document.write( "then \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( "50 mi accident travel time equation \n" ); document.write( " \n" ); document.write( "multiply by .6s to clear the denominators, results \n" ); document.write( ".6(50) + (d-50) = .6d + .6s(3) \n" ); document.write( "30 + d - 50 = .6d + 1.8s \n" ); document.write( "d - 20 = .6d + 1.8s \n" ); document.write( "d - .6d - 1.8s = 20 \n" ); document.write( ".4d - 1.8s = 20 \n" ); document.write( ": \n" ); document.write( "100 mi accident travel time equation \n" ); document.write( " \n" ); document.write( "multiply by .6s, results \n" ); document.write( ".6(100) + (d-100) = .6d + .6s(2) \n" ); document.write( "60 + d - 100 = .6d + 1.2s \n" ); document.write( "d - 40 = .6d + 1.2s \n" ); document.write( "d - .6d - 1.2s = 40 \n" ); document.write( ".4d - 1.2s = 40 \n" ); document.write( ": \n" ); document.write( "Subtract the 1st equation from the above equaton \n" ); document.write( ".4d - 1.2s = 40 \n" ); document.write( ".4d - 1.8s = 20 \n" ); document.write( "---------------- \n" ); document.write( "+.6s = 20 \n" ); document.write( "s = \n" ); document.write( "s = 33 \n" ); document.write( "Find d \n" ); document.write( ".4d - 1.2(33.33) = 40 \n" ); document.write( ".4d - 40 = 40 \n" ); document.write( ".4d = 40 + 40 \n" ); document.write( ".4d = 80 \n" ); document.write( "d = 80/.4 \n" ); document.write( "d = 200 mi is the distance \n" ); document.write( "; \n" ); document.write( ": \n" ); document.write( "See if the that checks out in the 100 mi accident equation \n" ); document.write( " \n" ); document.write( "3 + \n" ); document.write( "3 + 5 = 6 + 2 \n" ); document.write( ": \n" ); document.write( "We can say: normal speed = 33 \n" ); document.write( " \n" ); document.write( " |